Answer:
The probability that at least 280 of these students are smokers is 0.9664.
Step-by-step explanation:
Let the random variable <em>X</em> be defined as the number of students at a particular college who are smokers
The random variable <em>X</em> follows a Binomial distribution with parameters n = 500 and p = 0.60.
But the sample selected is too large and the probability of success is close to 0.50.
So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:
1. np ≥ 10
2. n(1 - p) ≥ 10
Check the conditions as follows:

Thus, a Normal approximation to binomial can be applied.
So,

Compute the probability that at least 280 of these students are smokers as follows:
Apply continuity correction:
P (X ≥ 280) = P (X > 280 + 0.50)
= P (X > 280.50)

*Use a <em>z</em>-table for the probability.
Thus, the probability that at least 280 of these students are smokers is 0.9664.
Answer:
D
Step-by-step explanation:
Answer:
You have 3.5 Servings.
Step-by-step explanation:
1/4 x 4 = 4/16
14/4 = 3.5
F ( x ) = 3 sim x + 3 cos x
f ` ( x ) = 3 cos x - 3 sin x
f `` ( x ) = - 3 sin x - 3 cos x = - 3 ( sin x + cos x )
The inflection points:
- 3 ( sin x + cos x ) = 0
sin x + cos x = 0
sin x = - cos x / : cos x
tan x = - 1
x 1 = 3π / 4
x 2 = 7π / 4
The function is concave up when f``(x) > 0
- 3( sin x+ cos x ) > 0
sin x + cos x < 0
tan x < - 1
f is concave up for:
x ∈ ( π/2, 3π/4 ) ∪ ( 3π/2, 7π/4)
f is concave down for:
x ∈ ( 0, π/2 ) ∪ ( 3π / 4, 3π/2 ) ∪ ( 7π / 4, 2 π ).